Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int fst[100005], nxt[200005], v[200005], tol;
int c[100005];
int fa[100005], sz[100005], dep[100005], son[100005];
int top[100005], scc[100005], _scc;
int rscc[100005];
vector<int> City[100005];
void edge(int a, int b) {
  v[tol] = b;
  nxt[tol] = fst[a];
  fst[a] = tol++;
}
void dfs1(int pa, int u, int deep) {
  dep[u] = deep;
  fa[u] = pa;
  sz[u] = 1;
  son[u] = -1;
  for (int e = fst[u]; ~e; e = nxt[e]) {
    if (v[e] == pa) continue;
    dfs1(u, v[e], deep + 1);
    sz[u] += sz[v[e]];
    if (son[u] == -1 || sz[son[u]] < sz[v[e]]) {
      son[u] = v[e];
    }
  }
}
void dfs2(int u) {
  scc[u] = (++_scc);
  rscc[_scc] = u;
  if (~son[u]) {
    top[son[u]] = top[u];
    dfs2(son[u]);
  }
  for (int e = fst[u]; ~e; e = nxt[e]) {
    if (v[e] == fa[u] || v[e] == son[u]) continue;
    top[v[e]] = v[e];
    dfs2(v[e]);
  }
}
vector<int> S[400005];
void merge(vector<int>& a, vector<int>& b) {
  vector<int> c;
  c.clear();
  int k = min(10, (int(a.size())) + (int(b.size())));
  int s = 0, t = 0;
  while (k--) {
    if (s == (int(a.size())))
      c.push_back(b[t++]);
    else if (t == (int(b.size())))
      c.push_back(a[s++]);
    else if (a[s] < b[t])
      c.push_back(a[s++]);
    else
      c.push_back(b[t++]);
  }
  a.clear();
  for (int i = int(0); i < int((int(c.size()))); i++) a.push_back(c[i]);
}
void build(int l, int r, int o) {
  S[o].clear();
  if (l == r) {
    merge(S[o], City[rscc[l]]);
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, (o << 1));
  build(mid + 1, r, ((o << 1) | 1));
  merge(S[o], S[(o << 1)]);
  merge(S[o], S[((o << 1) | 1)]);
}
vector<int> ans;
void query(int l, int r, int o, int L, int R) {
  if (L <= l && r <= R) {
    merge(ans, S[o]);
    return;
  }
  int mid = l + r >> 1;
  if (L <= mid) query(l, mid, (o << 1), L, R);
  if (mid < R) query(mid + 1, r, ((o << 1) | 1), L, R);
}
void LCA(int u, int v) {
  ans.clear();
  int f1 = top[u], f2 = top[v];
  while (f1 != f2) {
    if (dep[f1] < dep[f2]) {
      swap(u, v);
      swap(f1, f2);
    }
    query(1, _scc, 1, scc[f1], scc[u]);
    u = fa[f1];
    f1 = top[u];
  }
  if (dep[u] > dep[v]) {
    swap(u, v);
  }
  if (u) {
    query(1, _scc, 1, scc[u], scc[v]);
  } else if (scc[v] >= 2) {
    query(1, _scc, 1, 2, scc[v]);
  }
}
int main() {
  int a, b;
  tol = 0;
  memset(fst, -1, sizeof(fst));
  scanf("%d%d%d", &n, &m, &q);
  for (int i = int(0); i < int(n - 1); i++) {
    scanf("%d%d", &a, &b);
    edge(a, b);
    edge(b, a);
  }
  for (int i = int(1); i < int(m + 1); i++) {
    scanf("%d", c + i);
    City[c[i]].push_back(i);
  }
  dfs1(0, 1, 1);
  _scc = 0;
  top[1] = 1;
  dfs2(1);
  build(1, _scc, 1);
  int v, u;
  while (q--) {
    scanf("%d%d%d", &v, &u, &a);
    LCA(u, v);
    a = min(a, (int(ans.size())));
    printf("%d", a);
    for (int i = int(0); i < int(a); i++) printf(" %d", ans[i]);
    printf("\n");
  }
  return 0;
}